Given data

• Area of coil: $A = 70\,\text{cm}^2 = 70\times 10^{-4}\,\text{m}^2 = 7\times 10^{-3}\,\text{m}^2$
• Number of turns: $N = 600$
• Magnetic field: $B = 0.4\,\text{T}$
• Speed of rotation: $500\,\text{rev/min}$
• Plane of coil is inclined at $60^\circ$ with the magnetic field.

Angular speed

Frequency: $f = \frac{500}{60} = \frac{25}{3}\,\text{Hz}$

Therefore, $\omega = 2\pi f = 2\times \frac{22}{7}\times \frac{25}{3} = \frac{1100}{21}\,\text{rad/s}$

Instantaneous emf formula

For a rotating coil, $e = NBA\omega \sin\theta$ where $\theta$ is the angle between the normal to the coil and the magnetic field.

But the plane of the coil makes $60^\circ$ with the field, so the normal makes $\theta = 90^\circ - 60^\circ = 30^\circ$

Hence, $e = NBA\omega \sin 30^\circ$

Substitute values

First compute $NBA$: $NBA = 600\times 0.4\times 7\times 10^{-3}$ $= 600\times 0.0028 = 1.68$

Then, $e = 1.68\times \frac{1100}{21}\times \frac{1}{2}$

$e = 0.84\times \frac{1100}{21}$

$e = \frac{924}{21} = 44\,\text{V}$

Final answer

$\boxed{44}$

Comparison with stored answer

Stored correct answer = $44$

This matches the derived answer.